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INT

  • 3 years ago

change the cartesian integral into an equivalent polar integral:

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  1. INT
    • 3 years ago
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    \[\int\limits_{\sqrt{2}}^{2}\int\limits_{\sqrt{4-y^{2}}}^{y} dx dy\]

  2. amistre64
    • 3 years ago
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    |dw:1353881320734:dw|i think this is what it looks like at least

  3. amistre64
    • 3 years ago
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    remember that x=rcos(t) y=rsin(t) and r=sqrt(x^2+y^2), but thats really already given as 2

  4. INT
    • 3 years ago
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    I think I understand how to get pi/4 < theta < pi/2 But finding the bounds for r is throwing me off.

  5. amistre64
    • 3 years ago
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    well, since r only really changes from 0 to 2; the bounds for r are 0 to 2

  6. amistre64
    • 3 years ago
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    it r changed according to some function of t the r would be bound from 0 to f(t)

  7. INT
    • 3 years ago
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    according to the answers, r is bounded by 2 < r < 2csc(theta)

  8. amistre64
    • 3 years ago
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    |dw:1353881730467:dw|

  9. INT
    • 3 years ago
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    but at pi/2 r can be 2 and rad(2)

  10. amistre64
    • 3 years ago
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    it might help to see the limits in this fashion to be able to see how to change them out \[\Large \int_{y=\sqrt{2}}^{y=2}~~~\int_{x=\sqrt{4-y^{2}}}^{x=y} ~~~dx~dy\]

  11. INT
    • 3 years ago
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    my book says to plug in x=rcos(theta) etc and magically gets the polar bounds. They dont show any intermediate steps so im really confused.

  12. amistre64
    • 3 years ago
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    \[x= \sqrt{r^2-y^2}\]\[2cos(t) = \sqrt{2^2-(2sin(t))^2}\]\[4cos^2(t) = 4-4sin^2(t)\]\[cos^2(t) +sin^2(t)=1~\text{for any t}\] \[x= y\]\[2cos(t) = 2sin(t)\]\[t=\frac{pi}4\] something along those lines i believe

  13. amistre64
    • 3 years ago
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    y = 2 2sin(t) = 2 sin(t) = 1; when t=pi/2 y=sqrt(2) 2sin(t)=sqrt(2) sin(t)=sqrt(2)/2; when t=pi/4

  14. amistre64
    • 3 years ago
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    i might need to see the steps in your books to refresh me memory

  15. INT
    • 3 years ago
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    this is all that they show.

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  16. INT
    • 3 years ago
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    I just dont understand where they cot the csc from.

  17. amistre64
    • 3 years ago
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    your d parts in that png are dydx ... not dxdy

  18. INT
    • 3 years ago
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    well, thats a problem... the question has it as dx dy

  19. INT
    • 3 years ago
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    I dont think dy dx even works. Wouldnt it be y=y y=rad(4-y^2)

  20. amistre64
    • 3 years ago
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    yeah, so the png might be a typo

  21. INT
    • 3 years ago
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    ok so ignoring the png, would the r just be 0 to 2?

  22. amistre64
    • 3 years ago
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    from what i can tell yes; |dw:1353883154486:dw|

  23. amistre64
    • 3 years ago
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    but then i might have been thinking of this backwards

  24. amistre64
    • 3 years ago
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    so, r goes from 2 up to the line; its not the area inside the circle

  25. amistre64
    • 3 years ago
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    if i were to attempt this; I would determine the area of the triangle and subtract out the area inside the circle part

  26. INT
    • 3 years ago
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    But doesnt the outer circle ask that the Y bounds are 2 and rad(2) not the X?

  27. INT
    • 3 years ago
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    oh you flipped it

  28. amistre64
    • 3 years ago
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    i positioned the graph in a convenient manner; a swaped the x and y axis ,,, yes

  29. amistre64
    • 3 years ago
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    so, what we want is a function in ploar from for the line y=2

  30. INT
    • 3 years ago
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    well, even though the answer key messed up the dxdy, but 2-(pi/2) seems to be the right answer.

  31. amistre64
    • 3 years ago
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    it is; and youd be able to see that y=2 converts to 2csc(t) if i could remember how to do it :)

  32. amistre64
    • 3 years ago
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    got it r^2 = x^2 + y^2 ; but y=2 is constant r^2 = x^2 + 4 ; x = rcos(t) r^2 = r^2cos^2(t) + 4 r^2-r^2cos^2(t) = 4 r^2(1-cos^2(t)) = 4 r^2 sin^2(t) = 4 r^2 = 4csc^2(t) r = 2csc(t)

  33. amistre64
    • 3 years ago
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    area of triangle = 2*2/2 = 2 area of sector = 4pi/8 = pi/2 area of shaded region = triangle - sector: 2 - pi/2

  34. INT
    • 3 years ago
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    ok that makes sense.. one more question, is it a coincidence that y = 2 r sin(t) = 2 r = 2 / sin(t) using 1/sin(t) = csc(t) trig identity r = 2 csc(t) ?

  35. amistre64
    • 3 years ago
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    not sure if its coincidence or not; but that seems to work out as well :)

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